Uncorrelated random variables

Two square-integrable random variables $X,Y$ are uncorrelated random variables when $\mathrm{Cov}(X,Y)=0$. Independent random variables with finite second moments are uncorrelated random variables, but zero covariance does not generally imply independence.

For an explicit counterexample, let $X$ take $-1$, $0$, $1$ with equal probabilities, and let $Y=X^2$. Then $\mathbb{E}X=\mathbb{E}{X}^3=0$, so $\mathrm{Cov}(X,Y)=\mathbb{E}{X}^3-\mathbb{E}X\mathbb{E}{X}^2=0$. Nevertheless, $Y$ is completely determined by $X$: $\mathbb{P}(Y=0\mid X=0)=1$, while $\mathbb{P}(Y=0)=1/3$. Thus these are uncorrelated random variables with statistical dependence. In this example, opposite contributions from the odd function $x^3$ cancel; the even function $x^2$ still retains information about $|X|$. The last panel of the covariance diagram shows the relationship.

There are useful special cases. Uncorrelated jointly normal variables are independent, and the Bernoulli independence from zero covariance result supplies another setting where the converse holds. Their distribution assumptions do the extra work; zero covariance alone does not.

Here is an explicit joint distribution, for $0<p<0.8$: assign mass $(1+p)e^{-p}-p$ at $(0,0)$, mass $p(1-e^{-p})$ at $(1,0)$, mass $p{e}^{-p}$ at $(1,1)$, and mass $e^{-p}{p}^k/k!$ at $(0,k)$ for $k\ge 2$; all other masses are zero. These are nonnegative: the first is decreasing in $p$, and at $p=0.8$ it is still positive. Summing over rows and columns gives a Bernoulli distribution with mean $p$ and a Poisson distribution with mean $p$, so the joint mass sums to one.

Only $(1,1)$ contributes to the mixed first moment, giving $\mathbb{E}[XY]=p{e}^{-p}$ and the covariance above. The strictly decreasing function $e^{-p}-p$ has a unique zero in $(0,0.8)$, approximately $0.567143$. At that parameter the variables are uncorrelated. They are nevertheless dependent, because $\mathbb{P}(X=1\mid Y=1)=1$ whereas $\mathbb{P}(X=1)=p<1$. The Bernoulli independence from zero covariance result requires both variables to be Bernoulli and does not apply here.

For independent standard normal random variables ${\eta }_1,\dots ,{\eta }_n$, define the displayed errors for $i\ge 2$. Each has mean zero and variance one. Nonadjacent errors depend on disjoint sets of independent variables. Adjacent covariances vanish because their product contains an independent centered factor. Thus their covariance matrix is $I_n$.

They are nevertheless dependent: the Gaussian fourth moment gives $\mathbb{E}[{ϵ}_i^2{ϵ}_{i+1}^2]=3$, and hence $\mathrm{Cov}({ϵ}_i^2,{ϵ}_{i+1}^2)=2$. Constant variance and zero pairwise covariance can conceal dependence in magnitudes. This is useful when interpreting a residual plot in linear regression.

For two Bernoulli random variables with joint probabilities $a,b,c,d$ in the order $(0,0),(0,1),(1,0),(1,1)$, their covariance is $d-(c+d)(b+d)=ad-bc$. Zero covariance fixes the $(1,1)$ probability to the product of its marginals; subtracting this from each marginal fixes all three remaining joint probabilities to their products. Thus uncorrelated random variables taking only zero and one are independent, including degenerate marginals. The implication is special to this two-point setting and fails for general distributions.